Đáp án + Giải thích các bước giải:

$CuO+H_2\xrightarrow{t^o}Cu+H_2O(1)$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O(2)$

$Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O(3)$

$Fe+2HCl\to FeCl_2+H_2(4)$

$\to m_{Cu}=m_{rắn\,ko\,tan}=9,6(g)$

$\to n_{Cu}=\frac{9,6}{64}=0,15(mol)$

Theo PT: $n_{Fe(4)}=n_{H_2(4)}=\frac{15,68}{22,4}=0,7(mol)$

Theo PT: $n_{CuO}=n_{Cu}=0,15(mol)$

$\to m_{CuO}=0,15.80=12(g)$

$\to m_{Fe_2O_3}+m_{Fe_3O_4}=67,2-12=55,2(g)$

$\to 160n_{Fe_2O_3}+232n_{Fe_3O_4}=55,2(*)$

Theo PT: $2n_{Fe_2O_3}+3n_{Fe_3O_4}=n_{Fe(4)}=0,7(**)$

Từ $(*)(**)\to n_{Fe_2O_3}=0,2(mol);n_{Fe_3O_4}=0,1(mol)$

$\to m_{Fe_2O_3}=0,2.160=32(g)$

$\%m_{CuO}=\frac{12}{67,2}.100\%\approx 17,86\%$

$\%m_{Fe_2O_3}=\frac{32}{67,2}.100\%\approx 47,62\%$

$\%m_{Fe_3O_4}=100-17,86-47,62=34,52\%$